Since the cylinder is not rolling,
Let axis of rotation be at A
Torque, τ=Iα
Since α=0τ=0F×R+N1×0+μN1×0=μN2×R+N2×RF=(1+μ)N2⟶(1)
Balancing vertical forces
μN2+F+N1=mg⟶(2)
Balancing horizontal forces
μN1=N2⟶(3)
From 3
N1=N2μμN2+F+N2μ=mgN2=μmg−Fμμ2+1⟶(4)
From 4
F=(1+μ)(μmg−Fμ)μ2+1F=(1+μ)(1+μ)2μ2+μ+1F=μW.(1+μ)2μ2+μ+1